Question Number 99368 by 4635 last updated on 20/Jun/20 $${please}\:{sir}\:{my}\:{problem}\:{in}\:{my}\:{solution} \\ $$$${is}\:{where}? \\ $$ Commented by mr W last updated on 21/Jun/20 $${this}\:{is}\:{your}\:{first}\:{post}.\:{that}\:{means} \\ $$$${you}\:{didn}'{t}\:{post}\:{any}\:{question}\:{or}\:{any}…
Question Number 33800 by tawa tawa last updated on 25/Apr/18 Answered by ajfour last updated on 25/Apr/18 $${I}_{\mathrm{3}} ={I}_{{cell}} =\frac{\mathrm{6}{V}}{\left(\frac{\mathrm{5}×\mathrm{4}}{\mathrm{5}+\mathrm{4}}+\mathrm{3}+\mathrm{0}.\mathrm{5}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{6}}{\mathrm{5}.\mathrm{72}}\:\approx\:\mathrm{1}.\mathrm{05}{A} \\ $$$${I}_{\mathrm{5}} =\frac{\mathrm{4}}{\mathrm{9}}×\mathrm{1}.\mathrm{05}\:=\:\mathrm{0}.\mathrm{47}{A}…
Question Number 99314 by Dwaipayan Shikari last updated on 20/Jun/20 $${Find}\left[\right]{the}\left[\right]{value}\left[\right]{of} \\ $$$$\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}\sqrt{\mathrm{1}+\mathrm{6}\sqrt{\mathrm{1}+\mathrm{7}}}}}}}….\infty \\ $$ Commented by mr W last updated on 20/Jun/20 Commented by…
Question Number 33753 by 255ketto last updated on 23/Apr/18 Answered by MJS last updated on 23/Apr/18 $$\mathrm{now} \\ $$$$\mathrm{child}'\mathrm{s}\:\mathrm{age}\:{x} \\ $$$$\mathrm{mother}'\mathrm{s}\:\mathrm{age}\:\mathrm{4}{x} \\ $$$$\mathrm{five}\:\mathrm{years}\:\mathrm{ago} \\ $$$$\mathrm{child}'\mathrm{s}\:\mathrm{age}\:{x}−\mathrm{5}…
Question Number 164709 by Tawa11 last updated on 20/Jan/22 Answered by Tawa11 last updated on 20/Jan/22 $$\mathrm{Am}\:\mathrm{getting}\:\:\mathrm{14}.\mathrm{28}\:\:\mathrm{but}\:\mathrm{answer}\:\mathrm{at}\:\mathrm{the}\:\mathrm{back}\:\mathrm{is}\:\:\:\mathrm{12}.\mathrm{7}\:\:\:\:\mathrm{please}\:\mathrm{help}. \\ $$ Commented by mr W last updated…
Question Number 33575 by MURALI last updated on 19/Apr/18 $$\mathrm{simplify}\:\sqrt{\left(\mathrm{3}+\mathrm{2}\sqrt{\left.\mathrm{2}\right)}\right.} \\ $$ Answered by Rasheed.Sindhi last updated on 19/Apr/18 $$\mathrm{Let}\:\sqrt{\left(\mathrm{3}+\mathrm{2}\sqrt{\left.\mathrm{2}\right)}\right.}\:\:=\mathrm{a}+\mathrm{b}\sqrt{\mathrm{2}} \\ $$$$\mathrm{where}\:\mathrm{a},\mathrm{b}\:\in\mathbb{Z} \\ $$$$\:\:\:\left(\:\sqrt{\left(\mathrm{3}+\mathrm{2}\sqrt{\left.\mathrm{2}\right)}\right.}\:\right)^{\mathrm{2}} =\left(\mathrm{a}+\mathrm{b}\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}}…
Question Number 99097 by maths mind last updated on 18/Jun/20 $${Hello}\: \\ $$$${verry}\:{nice}\:{day}\:{for}\:{all}\:{of}\:{you}\:{god}\:{bless}\:{You} \\ $$$${pleas}\:{Can}\:{you}\:{use}\:{black}\:{Color}\:{shen}\:{You}\:{post}\:{Quation}\: \\ $$$${or}\:{Give}\:{answer}\:{is}\:{verry}\:{hard}\:{to}\:{read}\:{withe}\:{other}\:{colors} \\ $$ Commented by mr W last updated…
Question Number 33557 by JohnQuinceyStClair last updated on 19/Apr/18 $${This}^{} {is}^{} {a}^{} {formula}. \\ $$$${working}^{} {out}^{} {the}^{} {area}^{} {between}^{} {two} \\ $$$${Toroidal}^{} {coils}. \\ $$$${Both}^{}…
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Question Number 33508 by EFG last updated on 18/Apr/18 $$\frac{\mathrm{3}^{{x}} ×\mathrm{8}^{{x}} }{\mathrm{12}^{{x}+\mathrm{1}} } \\ $$ Commented by Rasheed.Sindhi last updated on 18/Apr/18 $$\left(\frac{\mathrm{3}×\mathrm{8}}{\mathrm{12}}\right)^{\mathrm{x}} ×\frac{\mathrm{1}}{\mathrm{12}}=\mathrm{2}^{\mathrm{x}} /\left(\mathrm{2}^{\mathrm{2}}…